Problem: Solve for $x$ : $ 6|x + 3| - 8 = 3|x + 3| + 6 $
Explanation: Subtract $ {3|x + 3|} $ from both sides: $ \begin{eqnarray} 6|x + 3| - 8 &=& 3|x + 3| + 6 \\ \\ { - 3|x + 3|} && { - 3|x + 3|} \\ \\ 3|x + 3| - 8 &=& 6 \end{eqnarray} $ Add ${8}$ to both sides: $ \begin{eqnarray} 3|x + 3| - 8 &=& 6 \\ \\ { + 8} &=& { + 8} \\ \\ 3|x + 3| &=& 14 \end{eqnarray} $ Divide both sides by ${3}$ $ \dfrac{3|x + 3|} {{3}} = \dfrac{14} {{3}} $ Simplify: $ |x + 3| = \dfrac{14}{3}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 3 = -\dfrac{14}{3} $ or $ x + 3 = \dfrac{14}{3} $ Solve for the solution where $x + 3$ is negative: $ x + 3 = -\dfrac{14}{3} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& -\dfrac{14}{3} \\ \\ {- 3} && {- 3} \\ \\ x &=& -\dfrac{14}{3} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $3$ $ x = - \dfrac{14}{3} {- \dfrac{9}{3}} $ $ x = -\dfrac{23}{3} $ Then calculate the solution where $x + 3$ is positive: $ x + 3 = \dfrac{14}{3} $ Subtract ${3}$ from both sides: $ \begin{eqnarray} x + 3 &=& \dfrac{14}{3} \\ \\ {- 3} && {- 3} \\ \\ x &=& \dfrac{14}{3} - 3 \end{eqnarray} $ Change the ${ - 3}$ to an equivalent fraction with a denominator of $3$ $ x = \dfrac{14}{3} {- \dfrac{9}{3}} $ $ x = \dfrac{5}{3} $ Thus, the correct answer is $x = -\dfrac{23}{3} $ or $x = \dfrac{5}{3} $.